Two discs are rotating about their axes, normal to the discs and passing through the centres of the

Question

Two discs are rotating about their axes, normal to the discs and passing through the centres of the discs. Disc $D_{1}$ has 2 kg mass and 0.2 m radius and initial angular velocity of $50 \mathrm{rad} \mathrm{s}^{-1}$. Disc $D_{2}$ has 4 kg mass, 0.1 m radius and initial angular velocity of $200 \mathrm{rad} \mathrm{s}^{-1}$. The two discs are brought in contact face to face, with their axes of rotation coincident. The final angular velocity (in $\mathrm{rad} \mathrm{s}^{-1}$ ) of the system is\

Professor PC Thomas Classes · Answer

(b) : Moment of inertia of disc $D_{1}$ about an axis passing through its centre and normal to its plane is $$ I_{1}=\frac{M R^{2}}{2}=\frac{(2 \mathrm{~kg})(0.2 \mathrm{~m})^{2}}{2}=0.04 \mathrm{~kg} \mathrm{~m}^{2} $$ Initial angular velocity of disc $D_{1}, \omega_{1}=50 \mathrm{rad} \mathrm{s}^{-1}$\ Moment of inertia of disc $D_{2}$ about an axis passing through its centre and normal to its plane is $$ I_{2}=\frac{(4 \mathrm{~kg})(0.1 \mathrm{~m})^{2}}{2}=0.02 \mathrm{~kg} \mathrm{~m}^{2} $$ Initial angular velocity of disc $D_{2}, \omega_{2}=200 \mathrm{rad} \mathrm{s}^{-1}$\ Total initial angular momentum of the two discs is $$ L_{i}=I_{1} \omega_{1}+I_{2} \omega_{2} $$ When two discs are brought in contact face to face (one on the top of the other) and their axes of rotation coincide, the moment of inertia $I$ of the system is equal to the sum of their\ individual moment of inertia. $$ I=I_{1}+I_{2} $$ Let $\omega$ be the final angular speed of the system. The final angular momentum of the system is $$ L_{f}=I \omega=\left(I_{1}+I_{2}ight) \omega $$ According to law of conservation of angular momentum, we get\ $L_{i}=L_{f}$ or, $I_{1} \omega_{1}+I_{2} \omega_{2}=\left(I_{1}+I_{2}ight) \omega$ or, $\omega=\frac{I_{1} \omega_{1}+I_{2} \omega_{2}}{I_{1}+I_{2}}$\ $=\frac{\left(0.04 \mathrm{~kg} \mathrm{~m}^{2}ight)\left(50 \mathrm{rad} \mathrm{s}^{-1}ight)+\left(0.02 \mathrm{~kg} \mathrm{~m}^{2}ight)\left(200 \mathrm{rad} \mathrm{s}^{-1}ight)}{(0.04+0.02) \mathrm{kg} \mathrm{m}^{2}}$\ $=\frac{(2+4)}{0.06} \mathrm{rad} \mathrm{s}^{-1}=100 \mathrm{rad} \mathrm{s}^{-1}$\